From: Logan Capaldo <logancapaldo@...>
Date: 2006-07-29T08:18:18+09:00
Subject: Re: Finding "\" in a string


On Jul 28, 2006, at 7:10 PM, Paul Battley wrote:

> On 28/07/06, Eric Armstrong <Eric.Armstrong@sun.com> wrote:
>> Very helpful. At least now I have a good idea that it
>> can be made to work. Now let's see if I can dial in my
>> understanding with respect to replacement strings...
>>
>> The fact that \1, \2 are recognized /shouldn't/ matter,
>> because the tokenizer really ought to ignore any \x
>> sequence where the x isn't a known special character.
>> But it's rapidly becoming apparent that the tokenizer
>> takes any \x and translates it to /something/, so we
>> need \\ to get one \ in the string.
>
> In fact, the tokenizer *does* ignore non-special sequences:
>
> "\\foo".gsub("\\", "\\x") # => "\\xfoo"
>
> And if there's nothing else left in the string, that works, too:
>
> "\\foo".gsub("\\", "\\") # => "\\foo"
>
> However, in order to be able to produce a literal \1, it also needs to
> understand \\. Therefore, this gives the same effect:
>
> "\\foo".gsub("\\", "\\\\") # => "\\foo"
>
> But I missed a trick earler. We only actually need six backslashes to
> emit two: the end of the string will terminate the sequence as in the
> second example:
>
> "\\foo".gsub("\\", "\\\\\\") # => "\\\\foo"
>
> Is it getting clearer?
>
> Paul.
>

This is why when I need to gsub backslashes I use a block:

gsub(%r{\}) { '\\' }